{"id":1163,"date":"2021-06-30T17:02:02","date_gmt":"2021-06-30T17:02:02","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/areas-of-regions-involving-curves\/"},"modified":"2021-11-17T01:57:41","modified_gmt":"2021-11-17T01:57:41","slug":"areas-of-regions-involving-curves","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/areas-of-regions-involving-curves\/","title":{"raw":"Areas of Regions Involving Curves","rendered":"Areas of Regions Involving Curves"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Determine the area of a region between two curves by integrating with respect to the independent variable<\/li>\r\n \t<li>Find the area of a compound region<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Area of a Region between Two Curves<\/h2>\r\n<p id=\"fs-id1167793369342\">Let [latex]f(x)[\/latex] and [latex]g(x)[\/latex] be continuous functions over an interval [latex]\\left[a,b\\right][\/latex] such that [latex]f(x)\\ge g(x)[\/latex] on [latex]\\left[a,b\\right].[\/latex] We want to find the area between the graphs of the functions, as shown in the following figure.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"225\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212609\/CNX_Calc_Figure_06_01_001.jpg\" alt=\"This figure is a graph in the first quadrant. There are two curves on the graph. The higher curve is labeled \u201cf(x)\u201d and the lower curve is labeled \u201cg(x)\u201d. There are two boundaries on the x-axis labeled a and b. There is shaded area between the two curves bounded by lines at x=a and x=b.\" width=\"225\" height=\"203\" \/> Figure 1. The area between the graphs of two functions, [latex]f(x)[\/latex] and [latex]g(x),[\/latex] on the interval [latex]\\left[a,b\\right].[\/latex][\/caption]\r\n<p id=\"fs-id1167794067547\">As we did before, we are going to partition the interval on the [latex]x\\text{-axis}[\/latex] and approximate the area between the graphs of the functions with rectangles. So, for [latex]i=0,1,2\\text{,\u2026},n,[\/latex] let [latex]P=\\left\\{{x}_{i}\\right\\}[\/latex] be a regular partition of [latex]\\left[a,b\\right].[\/latex] Then, for [latex]i=1,2\\text{,\u2026},n,[\/latex] choose a point [latex]{x}_{i}^{*}\\in \\left[{x}_{i-1},{x}_{i}\\right],[\/latex] and on each interval [latex]\\left[{x}_{i-1},{x}_{i}\\right][\/latex] construct a rectangle that extends vertically from [latex]g({x}_{i}^{*})[\/latex] to [latex]f({x}_{i}^{*}).[\/latex] Figure 2(a) shows the rectangles when [latex]{x}_{i}^{*}[\/latex] is selected to be the left endpoint of the interval and [latex]n=10.[\/latex] Figure 2(b) shows a representative rectangle in detail.<\/p>\r\n\r\n<div id=\"fs-id1167793366749\" class=\"textbox tryit\">\r\n<h3>Interactive<\/h3>\r\n<p id=\"fs-id1167794072750\"><a href=\"https:\/\/www.desmos.com\/calculator\/zgei5m3k2t\" target=\"_blank\" rel=\"noopener\">Use this calculator to learn more about the areas between two curves.<\/a><\/p>\r\n\r\n<\/div>\r\n[caption id=\"\" align=\"aligncenter\" width=\"422\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212611\/CNX_Calc_Figure_06_01_002.jpg\" alt=\"This figure has three graphs. The first graph has two curves, one over the other. In between the curves is a rectangle. The top of the rectangle is on the upper curve labeled \u201cf(x*)\u201d and the bottom of the rectangle is on the lower curve and labeled \u201cg(x*)\u201d. The second graph, labeled \u201c(a)\u201d, has two curves on the graph. The higher curve is labeled \u201cf(x)\u201d and the lower curve is labeled \u201cg(x)\u201d. There are two boundaries on the x-axis labeled a and b. There is shaded area between the two curves bounded by lines at x=a and x=b. The third graph, labeled \u201c(b)\u201d has two curves one over the other. The first curve is labeled \u201cf(x*)\u201d and the lower curve is labeled \u201cg(x*)\u201d. There is a shaded rectangle between the two. The width of the rectangle is labeled as \u201cdelta x\u201d.\" width=\"422\" height=\"267\" \/> Figure 2. (a)We can approximate the area between the graphs of two functions, [latex]f(x)[\/latex] and [latex]g(x),[\/latex] with rectangles. (b) The area of a typical rectangle goes from one curve to the other.[\/caption]\r\n<p id=\"fs-id1167793432313\">The height of each individual rectangle is [latex]f({x}_{i}^{*})-g({x}_{i}^{*})[\/latex] and the width of each rectangle is [latex]\\text{\u0394}x.[\/latex] Adding the areas of all the rectangles, we see that the area between the curves is approximated by<\/p>\r\n\r\n<div id=\"fs-id1167794207238\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A\\approx\\displaystyle\\sum_{i=1}^{n} \\left[f({x}_{i}^{*})-g({x}_{i}^{*})\\right]\\text{\u0394}x.[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167793386764\">This is a Riemann sum, so we take the limit as [latex]n\\to \\infty [\/latex] and we get<\/p>\r\n\r\n<div id=\"fs-id1167794160372\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A=\\underset{n\\to \\infty }{\\text{lim}}\\displaystyle\\sum_{i=1}^{n} \\left[f({x}_{i}^{*})-g({x}_{i}^{*})\\right]\\text{\u0394}x={\\displaystyle\\int }_{a}^{b}\\left[f(x)-g(x)\\right]dx.[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167794051267\">These findings are summarized in the following theorem.<\/p>\r\n\r\n<div id=\"fs-id1167793274150\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Finding the Area between Two Curves<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167793978678\">Let [latex]f(x)[\/latex] and [latex]g(x)[\/latex] be continuous functions such that [latex]f(x)\\ge g(x)[\/latex] over an interval [latex]\\left[a,b\\right].[\/latex] Let [latex]R[\/latex] denote the region bounded above by the graph of [latex]f(x),[\/latex] below by the graph of [latex]g(x),[\/latex] and on the left and right by the lines [latex]x=a[\/latex] and [latex]x=b,[\/latex] respectively. Then, the area of [latex]R[\/latex] is given by<\/p>\r\n\r\n<div id=\"fs-id1167794293252\" class=\"equation\" style=\"text-align: center;\">[latex]A={\\displaystyle\\int }_{a}^{b}\\left[f(x)-g(x)\\right]dx[\/latex]<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793870660\">We apply this theorem in the following example.<\/p>\r\n\r\n<div id=\"fs-id1167793900960\" class=\"textbook exercises\">\r\n<h3>Example: Finding the Area of a Region between Two Curves 1<\/h3>\r\nIf <em>R<\/em> is the region bounded above by the graph of the function [latex]f(x)=x+4[\/latex] and below by the graph of the function [latex]g(x)=3-\\frac{x}{2}[\/latex] over the interval [latex]\\left[1,4\\right],[\/latex] find the area of region [latex]R.[\/latex]\r\n<div id=\"fs-id1167793361764\" class=\"exercise\">[reveal-answer q=\"fs-id1167794055154\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794055154\"]\r\n<p id=\"fs-id1167794055154\">The region is depicted in the following figure.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"417\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212614\/CNX_Calc_Figure_06_01_003.jpg\" alt=\"This figure is has two linear graphs in the first quadrant. They are the functions f(x) = x+4 and g(x)= 3-x\/2. In between these lines is a shaded region, bounded above by f(x) and below by g(x). The shaded area is between x=1 and x=4.\" width=\"417\" height=\"422\" \/> Figure 3. A region between two curves is shown where one curve is always greater than the other.[\/caption]\r\n<p id=\"fs-id1167794284162\">We have<\/p>\r\n\r\n<div id=\"fs-id1167793372473\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill A&amp; ={\\displaystyle\\int }_{a}^{b}\\left[f(x)-g(x)\\right]dx\\hfill \\\\ &amp; ={\\displaystyle\\int }_{1}^{4}\\left[(x+4)-(3-\\frac{x}{2})\\right]dx={\\displaystyle\\int }_{1}^{4}\\left[\\frac{3x}{2}+1\\right]dx\\hfill \\\\ &amp; ={\\left[\\frac{3{x}^{2}}{4}+x\\right]|}_{1}^{4}=(16-\\frac{7}{4})=\\frac{57}{4}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1167793912500\">The area of the region is [latex]\\frac{57}{4}{\\text{units}}^{2}.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793957091\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nIf [latex]R[\/latex] is the region bounded by the graphs of the functions [latex]f(x)=\\frac{x}{2}+5[\/latex] and [latex]g(x)=x+\\frac{1}{2}[\/latex] over the interval [latex]\\left[1,5\\right],[\/latex] find the area of region [latex]R.[\/latex]\r\n<div>[reveal-answer q=\"568690\"]Hint[\/reveal-answer]<\/div>\r\n<div>[hidden-answer a=\"568690\"]Graph the functions to determine which function\u2019s graph forms the upper bound and which forms the lower bound, then follow the process used in the previous example.[\/hidden-answer]<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1167793940237\" class=\"exercise\">[reveal-answer q=\"fs-id1167793933114\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793933114\"]\r\n<p id=\"fs-id1167793933114\">12 units<sup>2<\/sup><\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1167793924679\">In the last example, we defined the interval of interest as part of the problem statement. Quite often, though, we want to define our interval of interest based on where the graphs of the two functions intersect. This is illustrated in the following example.<\/p>\r\n\r\n<div id=\"fs-id1167793852231\" class=\"textbook exercises\">\r\n<h3>Example: Finding the Area of a Region between Two Curves 2<\/h3>\r\nIf [latex]R[\/latex] is the region bounded above by the graph of the function [latex]f(x)=9-{(\\frac{x}{2})}^{2}[\/latex] and below by the graph of the function [latex]g(x)=6-x,[\/latex] find the area of region [latex]R.[\/latex]\r\n<div id=\"fs-id1167793479817\" class=\"exercise\">[reveal-answer q=\"fs-id1167794199953\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794199953\"]\r\n<p id=\"fs-id1167794199953\">The region is depicted in the following figure.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"357\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212617\/CNX_Calc_Figure_06_01_004.jpg\" alt=\"This figure is has two graphs in the first quadrant. They are the functions f(x) = 9-(x\/2)^2 and g(x)= 6-x. In between these graphs, an upside down parabola and a line, is a shaded region, bounded above by f(x) and below by g(x).\" width=\"357\" height=\"347\" \/> Figure 4. This graph shows the region below the graph of [latex]f(x)[\/latex] and above the graph of [latex]g(x).[\/latex][\/caption]\r\n<p id=\"fs-id1167793930934\">We first need to compute where the graphs of the functions intersect. Setting [latex]f(x)=g(x),[\/latex] we get<\/p>\r\n\r\n<div id=\"fs-id1167793607866\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill f(x)&amp; =\\hfill &amp; g(x)\\hfill \\\\ \\\\ \\hfill 9-{(\\frac{x}{2})}^{2}&amp; =\\hfill &amp; 6-x\\hfill \\\\ \\hfill 9-\\frac{{x}^{2}}{4}&amp; =\\hfill &amp; 6-x\\hfill \\\\ \\hfill 36-{x}^{2}&amp; =\\hfill &amp; 24-4x\\hfill \\\\ \\hfill {x}^{2}-4x-12&amp; =\\hfill &amp; 0\\hfill \\\\ \\hfill (x-6)(x+2)&amp; =\\hfill &amp; 0.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1167794061273\">The graphs of the functions intersect when [latex]x=6[\/latex] or [latex]x=-2,[\/latex] so we want to integrate from -2 to 6. Since [latex]f(x)\\ge g(x)[\/latex] for [latex]-2\\le x\\le 6,[\/latex] we obtain<\/p>\r\n\r\n<div id=\"fs-id1167794210350\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill A&amp; ={\\displaystyle\\int }_{a}^{b}\\left[f(x)-g(x)\\right]dx\\hfill \\\\ &amp; ={\\displaystyle\\int }_{-2}^{6}\\left[9-{(\\frac{x}{2})}^{2}-(6-x)\\right]dx={\\displaystyle\\int }_{-2}^{6}\\left[3-\\frac{{x}^{2}}{4}+x\\right]dx\\hfill \\\\ &amp; ={\\left[3x-\\frac{{x}^{3}}{12}+\\frac{{x}^{2}}{2}\\right]|}_{-2}^{6}=\\frac{64}{3}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1167793638870\">The area of the region is [latex]\\frac{64}{3}[\/latex] units<sup>2<\/sup>.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793470839\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nIf <em>R<\/em> is the region bounded above by the graph of the function [latex]f(x)=x[\/latex] and below by the graph of the function [latex]g(x)={x}^{4},[\/latex] find the area of region [latex]R.[\/latex]\r\n<div>[reveal-answer q=\"766176\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"766176\"]Use the process from the last example.[\/hidden-answer]<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1167794076151\" class=\"exercise\">[reveal-answer q=\"fs-id1167793276952\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793276952\"]\r\n<p id=\"fs-id1167793276952\">[latex]\\frac{3}{10}[\/latex] unit<sup>2<\/sup><\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/cm88bTFvRU4?controls=0&amp;start=616&amp;end=707&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.1AreaBetweenCurves616to707_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.1 Area Between Curves\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]5609[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Areas of Compound Regions<\/h2>\r\n<p id=\"fs-id1167793400867\">So far, we have required [latex]f(x)\\ge g(x)[\/latex] over the entire interval of interest, but what if we want to look at regions bounded by the graphs of functions that cross one another? In that case, we modify the process we just developed by using the absolute value function.<\/p>\r\n\r\n<div id=\"fs-id1167793655292\" class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Finding the Area of a Region between Curves That Cross<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167793589368\">Let [latex]f(x)[\/latex] and [latex]g(x)[\/latex] be continuous functions over an interval [latex]\\left[a,b\\right].[\/latex] Let [latex]R[\/latex] denote the region between the graphs of [latex]f(x)[\/latex] and [latex]g(x),[\/latex] and be bounded on the left and right by the lines [latex]x=a[\/latex] and [latex]x=b,[\/latex] respectively. Then, the area of [latex]R[\/latex] is given by<\/p>\r\n\r\n<div id=\"fs-id1167794162812\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A={\\displaystyle\\int }_{a}^{b}|f(x)-g(x)|dx[\/latex]<\/div>\r\n<\/div>\r\n<p id=\"fs-id1167793452414\">In practice, applying this theorem requires us to break up the interval [latex]\\left[a,b\\right][\/latex] and evaluate several integrals, depending on which of the function values is greater over a given part of the interval. We study this process in the following example.<\/p>\r\n\r\n<div id=\"fs-id1167793263971\" class=\"textbook exercises\">\r\n<h3>Example: Finding the Area of a Region Bounded by Functions That Cross<\/h3>\r\nIf <em>R<\/em> is the region between the graphs of the functions [latex]f(x)= \\sin x[\/latex] and [latex]g(x)= \\cos x[\/latex] over the interval [latex]\\left[0,\\pi \\right],[\/latex] find the area of region [latex]R.[\/latex]\r\n<div id=\"fs-id1167793263974\" class=\"exercise\">[reveal-answer q=\"fs-id1167793510078\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793510078\"]\r\n<p id=\"fs-id1167793510078\">The region is depicted in the following figure.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"348\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212620\/CNX_Calc_Figure_06_01_005.jpg\" alt=\"This figure is has two graphs. They are the functions f(x) = sinx and g(x)= cosx. They are both periodic functions that resemble waves. There are two shaded areas between the graphs. The first shaded area is labeled \u201cR1\u201d and has g(x) above f(x). This region begins at the y-axis and stops where the curves intersect. The second region is labeled \u201cR2\u201d and begins at the intersection with f(x) above g(x). The shaded region stops at x=pi.\" width=\"348\" height=\"347\" \/> Figure 5. The region between two curves can be broken into two sub-regions.[\/caption]\r\n<p id=\"fs-id1167793420158\">The graphs of the functions intersect at [latex]x=\\pi \\text{\/}4.[\/latex] For [latex]x\\in \\left[0,\\pi \\text{\/}4\\right],[\/latex] [latex] \\cos x\\ge \\sin x,[\/latex] so<\/p>\r\n\r\n<div id=\"fs-id1167794325185\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]|f(x)-g(x)|=| \\sin x- \\cos x|= \\cos x- \\sin x[\/latex]<\/div>\r\n<p id=\"fs-id1167793258792\">On the other hand, for [latex]x\\in \\left[\\pi \\text{\/}4,\\pi \\right],[\/latex] [latex] \\sin x\\ge \\cos x,[\/latex] so<\/p>\r\n\r\n<div id=\"fs-id1167794026884\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]|f(x)-g(x)|=| \\sin x- \\cos x|= \\sin x- \\cos x[\/latex]<\/div>\r\n<p id=\"fs-id1167794040499\">Then<\/p>\r\n\r\n<div id=\"fs-id1167794040502\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill A&amp; ={\\displaystyle\\int }_{a}^{b}|f(x)-g(x)|dx\\hfill \\\\ &amp; ={\\displaystyle\\int }_{0}^{\\pi }| \\sin x- \\cos x|dx={\\displaystyle\\int }_{0}^{\\pi \\text{\/}4}( \\cos x- \\sin x)dx+{\\displaystyle\\int }_{\\pi \\text{\/}4}^{\\pi }( \\sin x- \\cos x)dx\\hfill \\\\ &amp; ={\\left[ \\sin x+ \\cos x\\right]|}_{0}^{\\pi \\text{\/}4}+{\\left[\\text{\u2212} \\cos x- \\sin x\\right]|}_{\\pi \\text{\/}4}^{\\pi }\\hfill \\\\ &amp; =(\\sqrt{2}-1)+(1+\\sqrt{2})=2\\sqrt{2}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1167794117704\">The area of the region is [latex]2\\sqrt{2}[\/latex] units<sup>2<\/sup>.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167793278122\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nIf <em>R<\/em> is the region between the graphs of the functions [latex]f(x)= \\sin x[\/latex] and [latex]g(x)= \\cos x[\/latex] over the interval [latex]\\left[\\frac{\\pi}{2},2\\pi \\right],[\/latex] find the area of region [latex]R.[\/latex]\r\n<div>[reveal-answer q=\"423799\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"423799\"]The two curves intersect at [latex]x=\\frac{(5\\pi )}{4}.[\/latex][\/hidden-answer]<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1167794121633\" class=\"exercise\">[reveal-answer q=\"fs-id1167793932268\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793932268\"]\r\n<p id=\"fs-id1167793932268\">[latex]2+2\\sqrt{2}[\/latex] units<sup>2<\/sup><\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/cm88bTFvRU4?controls=0&amp;start=1007&amp;end=1168&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.1AreaBetweenCurves1007to1168_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.1 Area Between Curves\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1167793455088\" class=\"textbook exercises\">\r\n<h3>Example: Finding the Area of a Complex Region<\/h3>\r\n<p id=\"fs-id1167793517455\">Consider the region depicted in Figure 6. Find the area of [latex]R.[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"229\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212623\/CNX_Calc_Figure_06_01_006.jpg\" alt=\"This figure is has two graphs in the first quadrant. They are the functions f(x) = x^2 and g(x)= 2-x. In between these graphs is a shaded region, bounded to the left by f(x) and to the right by g(x). All of which is above the x-axis. The region is labeled R. The shaded area is between x=0 and x=2.\" width=\"229\" height=\"242\" \/> Figure 6. Two integrals are required to calculate the area of this region.[\/caption]\r\n\r\n[reveal-answer q=\"fs-id1167793473492\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793473492\"]\r\n<p id=\"fs-id1167793473492\">As with the last example, we need to divide the interval into two pieces. The graphs of the functions intersect at [latex]x=1[\/latex] (set [latex]f(x)=g(x)[\/latex] and solve for [latex]x[\/latex]), so we evaluate two separate integrals: one over the interval [latex]\\left[0,1\\right][\/latex] and one over the interval [latex]\\left[1,2\\right].[\/latex]<\/p>\r\n<p id=\"fs-id1167793615262\">Over the interval [latex]\\left[0,1\\right],[\/latex] the region is bounded above by [latex]f(x)={x}^{2}[\/latex] and below by the [latex]x[\/latex]-axis, so we have<\/p>\r\n\r\n<div id=\"fs-id1167793564130\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{A}_{1}={\\displaystyle\\int }_{0}^{1}{x}^{2}dx={\\frac{{x}^{3}}{3}|}_{0}^{1}=\\frac{1}{3}.[\/latex]<\/div>\r\n<p id=\"fs-id1167793276900\">Over the interval [latex]\\left[1,2\\right],[\/latex] the region is bounded above by [latex]g(x)=2-x[\/latex] and below by the [latex]x\\text{-axis,}[\/latex] so we have<\/p>\r\n\r\n<div id=\"fs-id1167793372769\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{A}_{2}={\\displaystyle\\int }_{1}^{2}(2-x)dx={\\left[2x-\\frac{{x}^{2}}{2}\\right]|}_{1}^{2}=\\frac{1}{2}.[\/latex]<\/div>\r\n<p id=\"fs-id1167793619930\">Adding these areas together, we obtain<\/p>\r\n\r\n<div id=\"fs-id1167793619934\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A={A}_{1}+{A}_{2}=\\frac{1}{3}+\\frac{1}{2}=\\frac{5}{6}.[\/latex]<\/div>\r\n<p id=\"fs-id1167794209440\">The area of the region is [latex]\\frac{5}{6}[\/latex] units<sup>2<\/sup>.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1167793456255\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1167793581456\">Consider the region depicted in the following figure. Find the area of [latex]R.[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"250\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212625\/CNX_Calc_Figure_06_01_007.jpg\" alt=\"This figure is has two graphs in the first quadrant. They are the functions f(x) = squareroot of x and g(x)= 3\/2 \u2013 x\/2. In between these graphs is a shaded region, bounded to the left by f(x) and to the right by g(x). All of which is above the x-axis. The shaded area is between x=0 and x=3.\" width=\"250\" height=\"235\" \/> Figure 7.[\/caption]\r\n\r\n[reveal-answer q=\"217550\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"217550\"]The two curves intersect at [latex]x=1.[\/latex][\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1167793262779\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793262779\"]\r\n<p id=\"fs-id1167793262779\">[latex]\\frac{5}{3}[\/latex] units<sup>2<\/sup><\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try It.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/cm88bTFvRU4?controls=0&amp;start=1269&amp;end=1353&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266835\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266835\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.1AreaBetweenCurves1269to1353_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.1 Area Between Curves\" here (opens in new window)<\/a>.[\/hidden-answer]","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Determine the area of a region between two curves by integrating with respect to the independent variable<\/li>\n<li>Find the area of a compound region<\/li>\n<\/ul>\n<\/div>\n<h2>Area of a Region between Two Curves<\/h2>\n<p id=\"fs-id1167793369342\">Let [latex]f(x)[\/latex] and [latex]g(x)[\/latex] be continuous functions over an interval [latex]\\left[a,b\\right][\/latex] such that [latex]f(x)\\ge g(x)[\/latex] on [latex]\\left[a,b\\right].[\/latex] We want to find the area between the graphs of the functions, as shown in the following figure.<\/p>\n<div style=\"width: 235px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212609\/CNX_Calc_Figure_06_01_001.jpg\" alt=\"This figure is a graph in the first quadrant. There are two curves on the graph. The higher curve is labeled \u201cf(x)\u201d and the lower curve is labeled \u201cg(x)\u201d. There are two boundaries on the x-axis labeled a and b. There is shaded area between the two curves bounded by lines at x=a and x=b.\" width=\"225\" height=\"203\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. The area between the graphs of two functions, [latex]f(x)[\/latex] and [latex]g(x),[\/latex] on the interval [latex]\\left[a,b\\right].[\/latex]<\/p>\n<\/div>\n<p id=\"fs-id1167794067547\">As we did before, we are going to partition the interval on the [latex]x\\text{-axis}[\/latex] and approximate the area between the graphs of the functions with rectangles. So, for [latex]i=0,1,2\\text{,\u2026},n,[\/latex] let [latex]P=\\left\\{{x}_{i}\\right\\}[\/latex] be a regular partition of [latex]\\left[a,b\\right].[\/latex] Then, for [latex]i=1,2\\text{,\u2026},n,[\/latex] choose a point [latex]{x}_{i}^{*}\\in \\left[{x}_{i-1},{x}_{i}\\right],[\/latex] and on each interval [latex]\\left[{x}_{i-1},{x}_{i}\\right][\/latex] construct a rectangle that extends vertically from [latex]g({x}_{i}^{*})[\/latex] to [latex]f({x}_{i}^{*}).[\/latex] Figure 2(a) shows the rectangles when [latex]{x}_{i}^{*}[\/latex] is selected to be the left endpoint of the interval and [latex]n=10.[\/latex] Figure 2(b) shows a representative rectangle in detail.<\/p>\n<div id=\"fs-id1167793366749\" class=\"textbox tryit\">\n<h3>Interactive<\/h3>\n<p id=\"fs-id1167794072750\"><a href=\"https:\/\/www.desmos.com\/calculator\/zgei5m3k2t\" target=\"_blank\" rel=\"noopener\">Use this calculator to learn more about the areas between two curves.<\/a><\/p>\n<\/div>\n<div style=\"width: 432px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212611\/CNX_Calc_Figure_06_01_002.jpg\" alt=\"This figure has three graphs. The first graph has two curves, one over the other. In between the curves is a rectangle. The top of the rectangle is on the upper curve labeled \u201cf(x*)\u201d and the bottom of the rectangle is on the lower curve and labeled \u201cg(x*)\u201d. The second graph, labeled \u201c(a)\u201d, has two curves on the graph. The higher curve is labeled \u201cf(x)\u201d and the lower curve is labeled \u201cg(x)\u201d. There are two boundaries on the x-axis labeled a and b. There is shaded area between the two curves bounded by lines at x=a and x=b. The third graph, labeled \u201c(b)\u201d has two curves one over the other. The first curve is labeled \u201cf(x*)\u201d and the lower curve is labeled \u201cg(x*)\u201d. There is a shaded rectangle between the two. The width of the rectangle is labeled as \u201cdelta x\u201d.\" width=\"422\" height=\"267\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. (a)We can approximate the area between the graphs of two functions, [latex]f(x)[\/latex] and [latex]g(x),[\/latex] with rectangles. (b) The area of a typical rectangle goes from one curve to the other.<\/p>\n<\/div>\n<p id=\"fs-id1167793432313\">The height of each individual rectangle is [latex]f({x}_{i}^{*})-g({x}_{i}^{*})[\/latex] and the width of each rectangle is [latex]\\text{\u0394}x.[\/latex] Adding the areas of all the rectangles, we see that the area between the curves is approximated by<\/p>\n<div id=\"fs-id1167794207238\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A\\approx\\displaystyle\\sum_{i=1}^{n} \\left[f({x}_{i}^{*})-g({x}_{i}^{*})\\right]\\text{\u0394}x.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167793386764\">This is a Riemann sum, so we take the limit as [latex]n\\to \\infty[\/latex] and we get<\/p>\n<div id=\"fs-id1167794160372\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A=\\underset{n\\to \\infty }{\\text{lim}}\\displaystyle\\sum_{i=1}^{n} \\left[f({x}_{i}^{*})-g({x}_{i}^{*})\\right]\\text{\u0394}x={\\displaystyle\\int }_{a}^{b}\\left[f(x)-g(x)\\right]dx.[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167794051267\">These findings are summarized in the following theorem.<\/p>\n<div id=\"fs-id1167793274150\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Finding the Area between Two Curves<\/h3>\n<hr \/>\n<p id=\"fs-id1167793978678\">Let [latex]f(x)[\/latex] and [latex]g(x)[\/latex] be continuous functions such that [latex]f(x)\\ge g(x)[\/latex] over an interval [latex]\\left[a,b\\right].[\/latex] Let [latex]R[\/latex] denote the region bounded above by the graph of [latex]f(x),[\/latex] below by the graph of [latex]g(x),[\/latex] and on the left and right by the lines [latex]x=a[\/latex] and [latex]x=b,[\/latex] respectively. Then, the area of [latex]R[\/latex] is given by<\/p>\n<div id=\"fs-id1167794293252\" class=\"equation\" style=\"text-align: center;\">[latex]A={\\displaystyle\\int }_{a}^{b}\\left[f(x)-g(x)\\right]dx[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<p id=\"fs-id1167793870660\">We apply this theorem in the following example.<\/p>\n<div id=\"fs-id1167793900960\" class=\"textbook exercises\">\n<h3>Example: Finding the Area of a Region between Two Curves 1<\/h3>\n<p>If <em>R<\/em> is the region bounded above by the graph of the function [latex]f(x)=x+4[\/latex] and below by the graph of the function [latex]g(x)=3-\\frac{x}{2}[\/latex] over the interval [latex]\\left[1,4\\right],[\/latex] find the area of region [latex]R.[\/latex]<\/p>\n<div id=\"fs-id1167793361764\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794055154\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167794055154\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794055154\">The region is depicted in the following figure.<\/p>\n<div style=\"width: 427px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212614\/CNX_Calc_Figure_06_01_003.jpg\" alt=\"This figure is has two linear graphs in the first quadrant. They are the functions f(x) = x+4 and g(x)= 3-x\/2. In between these lines is a shaded region, bounded above by f(x) and below by g(x). The shaded area is between x=1 and x=4.\" width=\"417\" height=\"422\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. A region between two curves is shown where one curve is always greater than the other.<\/p>\n<\/div>\n<p id=\"fs-id1167794284162\">We have<\/p>\n<div id=\"fs-id1167793372473\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill A& ={\\displaystyle\\int }_{a}^{b}\\left[f(x)-g(x)\\right]dx\\hfill \\\\ & ={\\displaystyle\\int }_{1}^{4}\\left[(x+4)-(3-\\frac{x}{2})\\right]dx={\\displaystyle\\int }_{1}^{4}\\left[\\frac{3x}{2}+1\\right]dx\\hfill \\\\ & ={\\left[\\frac{3{x}^{2}}{4}+x\\right]|}_{1}^{4}=(16-\\frac{7}{4})=\\frac{57}{4}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793912500\">The area of the region is [latex]\\frac{57}{4}{\\text{units}}^{2}.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793957091\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>If [latex]R[\/latex] is the region bounded by the graphs of the functions [latex]f(x)=\\frac{x}{2}+5[\/latex] and [latex]g(x)=x+\\frac{1}{2}[\/latex] over the interval [latex]\\left[1,5\\right],[\/latex] find the area of region [latex]R.[\/latex]<\/p>\n<div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q568690\">Hint<\/span><\/div>\n<div>\n<div id=\"q568690\" class=\"hidden-answer\" style=\"display: none\">Graph the functions to determine which function\u2019s graph forms the upper bound and which forms the lower bound, then follow the process used in the previous example.<\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<div id=\"fs-id1167793940237\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793933114\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793933114\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793933114\">12 units<sup>2<\/sup><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167793924679\">In the last example, we defined the interval of interest as part of the problem statement. Quite often, though, we want to define our interval of interest based on where the graphs of the two functions intersect. This is illustrated in the following example.<\/p>\n<div id=\"fs-id1167793852231\" class=\"textbook exercises\">\n<h3>Example: Finding the Area of a Region between Two Curves 2<\/h3>\n<p>If [latex]R[\/latex] is the region bounded above by the graph of the function [latex]f(x)=9-{(\\frac{x}{2})}^{2}[\/latex] and below by the graph of the function [latex]g(x)=6-x,[\/latex] find the area of region [latex]R.[\/latex]<\/p>\n<div id=\"fs-id1167793479817\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167794199953\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167794199953\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794199953\">The region is depicted in the following figure.<\/p>\n<div style=\"width: 367px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212617\/CNX_Calc_Figure_06_01_004.jpg\" alt=\"This figure is has two graphs in the first quadrant. They are the functions f(x) = 9-(x\/2)^2 and g(x)= 6-x. In between these graphs, an upside down parabola and a line, is a shaded region, bounded above by f(x) and below by g(x).\" width=\"357\" height=\"347\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4. This graph shows the region below the graph of [latex]f(x)[\/latex] and above the graph of [latex]g(x).[\/latex]<\/p>\n<\/div>\n<p id=\"fs-id1167793930934\">We first need to compute where the graphs of the functions intersect. Setting [latex]f(x)=g(x),[\/latex] we get<\/p>\n<div id=\"fs-id1167793607866\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ccc}\\hfill f(x)& =\\hfill & g(x)\\hfill \\\\ \\\\ \\hfill 9-{(\\frac{x}{2})}^{2}& =\\hfill & 6-x\\hfill \\\\ \\hfill 9-\\frac{{x}^{2}}{4}& =\\hfill & 6-x\\hfill \\\\ \\hfill 36-{x}^{2}& =\\hfill & 24-4x\\hfill \\\\ \\hfill {x}^{2}-4x-12& =\\hfill & 0\\hfill \\\\ \\hfill (x-6)(x+2)& =\\hfill & 0.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167794061273\">The graphs of the functions intersect when [latex]x=6[\/latex] or [latex]x=-2,[\/latex] so we want to integrate from -2 to 6. Since [latex]f(x)\\ge g(x)[\/latex] for [latex]-2\\le x\\le 6,[\/latex] we obtain<\/p>\n<div id=\"fs-id1167794210350\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill A& ={\\displaystyle\\int }_{a}^{b}\\left[f(x)-g(x)\\right]dx\\hfill \\\\ & ={\\displaystyle\\int }_{-2}^{6}\\left[9-{(\\frac{x}{2})}^{2}-(6-x)\\right]dx={\\displaystyle\\int }_{-2}^{6}\\left[3-\\frac{{x}^{2}}{4}+x\\right]dx\\hfill \\\\ & ={\\left[3x-\\frac{{x}^{3}}{12}+\\frac{{x}^{2}}{2}\\right]|}_{-2}^{6}=\\frac{64}{3}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167793638870\">The area of the region is [latex]\\frac{64}{3}[\/latex] units<sup>2<\/sup>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793470839\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>If <em>R<\/em> is the region bounded above by the graph of the function [latex]f(x)=x[\/latex] and below by the graph of the function [latex]g(x)={x}^{4},[\/latex] find the area of region [latex]R.[\/latex]<\/p>\n<div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q766176\">Hint<\/span><\/p>\n<div id=\"q766176\" class=\"hidden-answer\" style=\"display: none\">Use the process from the last example.<\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<div id=\"fs-id1167794076151\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793276952\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793276952\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793276952\">[latex]\\frac{3}{10}[\/latex] unit<sup>2<\/sup><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/cm88bTFvRU4?controls=0&amp;start=616&amp;end=707&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.1AreaBetweenCurves616to707_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.1 Area Between Curves&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm5609\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5609&theme=oea&iframe_resize_id=ohm5609&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Areas of Compound Regions<\/h2>\n<p id=\"fs-id1167793400867\">So far, we have required [latex]f(x)\\ge g(x)[\/latex] over the entire interval of interest, but what if we want to look at regions bounded by the graphs of functions that cross one another? In that case, we modify the process we just developed by using the absolute value function.<\/p>\n<div id=\"fs-id1167793655292\" class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Finding the Area of a Region between Curves That Cross<\/h3>\n<hr \/>\n<p id=\"fs-id1167793589368\">Let [latex]f(x)[\/latex] and [latex]g(x)[\/latex] be continuous functions over an interval [latex]\\left[a,b\\right].[\/latex] Let [latex]R[\/latex] denote the region between the graphs of [latex]f(x)[\/latex] and [latex]g(x),[\/latex] and be bounded on the left and right by the lines [latex]x=a[\/latex] and [latex]x=b,[\/latex] respectively. Then, the area of [latex]R[\/latex] is given by<\/p>\n<div id=\"fs-id1167794162812\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A={\\displaystyle\\int }_{a}^{b}|f(x)-g(x)|dx[\/latex]<\/div>\n<\/div>\n<p id=\"fs-id1167793452414\">In practice, applying this theorem requires us to break up the interval [latex]\\left[a,b\\right][\/latex] and evaluate several integrals, depending on which of the function values is greater over a given part of the interval. We study this process in the following example.<\/p>\n<div id=\"fs-id1167793263971\" class=\"textbook exercises\">\n<h3>Example: Finding the Area of a Region Bounded by Functions That Cross<\/h3>\n<p>If <em>R<\/em> is the region between the graphs of the functions [latex]f(x)= \\sin x[\/latex] and [latex]g(x)= \\cos x[\/latex] over the interval [latex]\\left[0,\\pi \\right],[\/latex] find the area of region [latex]R.[\/latex]<\/p>\n<div id=\"fs-id1167793263974\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793510078\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793510078\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793510078\">The region is depicted in the following figure.<\/p>\n<div style=\"width: 358px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212620\/CNX_Calc_Figure_06_01_005.jpg\" alt=\"This figure is has two graphs. They are the functions f(x) = sinx and g(x)= cosx. They are both periodic functions that resemble waves. There are two shaded areas between the graphs. The first shaded area is labeled \u201cR1\u201d and has g(x) above f(x). This region begins at the y-axis and stops where the curves intersect. The second region is labeled \u201cR2\u201d and begins at the intersection with f(x) above g(x). The shaded region stops at x=pi.\" width=\"348\" height=\"347\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 5. The region between two curves can be broken into two sub-regions.<\/p>\n<\/div>\n<p id=\"fs-id1167793420158\">The graphs of the functions intersect at [latex]x=\\pi \\text{\/}4.[\/latex] For [latex]x\\in \\left[0,\\pi \\text{\/}4\\right],[\/latex] [latex]\\cos x\\ge \\sin x,[\/latex] so<\/p>\n<div id=\"fs-id1167794325185\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]|f(x)-g(x)|=| \\sin x- \\cos x|= \\cos x- \\sin x[\/latex]<\/div>\n<p id=\"fs-id1167793258792\">On the other hand, for [latex]x\\in \\left[\\pi \\text{\/}4,\\pi \\right],[\/latex] [latex]\\sin x\\ge \\cos x,[\/latex] so<\/p>\n<div id=\"fs-id1167794026884\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]|f(x)-g(x)|=| \\sin x- \\cos x|= \\sin x- \\cos x[\/latex]<\/div>\n<p id=\"fs-id1167794040499\">Then<\/p>\n<div id=\"fs-id1167794040502\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill A& ={\\displaystyle\\int }_{a}^{b}|f(x)-g(x)|dx\\hfill \\\\ & ={\\displaystyle\\int }_{0}^{\\pi }| \\sin x- \\cos x|dx={\\displaystyle\\int }_{0}^{\\pi \\text{\/}4}( \\cos x- \\sin x)dx+{\\displaystyle\\int }_{\\pi \\text{\/}4}^{\\pi }( \\sin x- \\cos x)dx\\hfill \\\\ & ={\\left[ \\sin x+ \\cos x\\right]|}_{0}^{\\pi \\text{\/}4}+{\\left[\\text{\u2212} \\cos x- \\sin x\\right]|}_{\\pi \\text{\/}4}^{\\pi }\\hfill \\\\ & =(\\sqrt{2}-1)+(1+\\sqrt{2})=2\\sqrt{2}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167794117704\">The area of the region is [latex]2\\sqrt{2}[\/latex] units<sup>2<\/sup>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793278122\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>If <em>R<\/em> is the region between the graphs of the functions [latex]f(x)= \\sin x[\/latex] and [latex]g(x)= \\cos x[\/latex] over the interval [latex]\\left[\\frac{\\pi}{2},2\\pi \\right],[\/latex] find the area of region [latex]R.[\/latex]<\/p>\n<div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q423799\">Hint<\/span><\/p>\n<div id=\"q423799\" class=\"hidden-answer\" style=\"display: none\">The two curves intersect at [latex]x=\\frac{(5\\pi )}{4}.[\/latex]<\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<div id=\"fs-id1167794121633\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793932268\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793932268\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793932268\">[latex]2+2\\sqrt{2}[\/latex] units<sup>2<\/sup><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/cm88bTFvRU4?controls=0&amp;start=1007&amp;end=1168&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.1AreaBetweenCurves1007to1168_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.1 Area Between Curves&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1167793455088\" class=\"textbook exercises\">\n<h3>Example: Finding the Area of a Complex Region<\/h3>\n<p id=\"fs-id1167793517455\">Consider the region depicted in Figure 6. Find the area of [latex]R.[\/latex]<\/p>\n<div style=\"width: 239px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212623\/CNX_Calc_Figure_06_01_006.jpg\" alt=\"This figure is has two graphs in the first quadrant. They are the functions f(x) = x^2 and g(x)= 2-x. In between these graphs is a shaded region, bounded to the left by f(x) and to the right by g(x). All of which is above the x-axis. The region is labeled R. The shaded area is between x=0 and x=2.\" width=\"229\" height=\"242\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 6. Two integrals are required to calculate the area of this region.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793473492\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793473492\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793473492\">As with the last example, we need to divide the interval into two pieces. The graphs of the functions intersect at [latex]x=1[\/latex] (set [latex]f(x)=g(x)[\/latex] and solve for [latex]x[\/latex]), so we evaluate two separate integrals: one over the interval [latex]\\left[0,1\\right][\/latex] and one over the interval [latex]\\left[1,2\\right].[\/latex]<\/p>\n<p id=\"fs-id1167793615262\">Over the interval [latex]\\left[0,1\\right],[\/latex] the region is bounded above by [latex]f(x)={x}^{2}[\/latex] and below by the [latex]x[\/latex]-axis, so we have<\/p>\n<div id=\"fs-id1167793564130\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{A}_{1}={\\displaystyle\\int }_{0}^{1}{x}^{2}dx={\\frac{{x}^{3}}{3}|}_{0}^{1}=\\frac{1}{3}.[\/latex]<\/div>\n<p id=\"fs-id1167793276900\">Over the interval [latex]\\left[1,2\\right],[\/latex] the region is bounded above by [latex]g(x)=2-x[\/latex] and below by the [latex]x\\text{-axis,}[\/latex] so we have<\/p>\n<div id=\"fs-id1167793372769\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{A}_{2}={\\displaystyle\\int }_{1}^{2}(2-x)dx={\\left[2x-\\frac{{x}^{2}}{2}\\right]|}_{1}^{2}=\\frac{1}{2}.[\/latex]<\/div>\n<p id=\"fs-id1167793619930\">Adding these areas together, we obtain<\/p>\n<div id=\"fs-id1167793619934\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A={A}_{1}+{A}_{2}=\\frac{1}{3}+\\frac{1}{2}=\\frac{5}{6}.[\/latex]<\/div>\n<p id=\"fs-id1167794209440\">The area of the region is [latex]\\frac{5}{6}[\/latex] units<sup>2<\/sup>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167793456255\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1167793581456\">Consider the region depicted in the following figure. Find the area of [latex]R.[\/latex]<\/p>\n<div style=\"width: 260px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11212625\/CNX_Calc_Figure_06_01_007.jpg\" alt=\"This figure is has two graphs in the first quadrant. They are the functions f(x) = squareroot of x and g(x)= 3\/2 \u2013 x\/2. In between these graphs is a shaded region, bounded to the left by f(x) and to the right by g(x). All of which is above the x-axis. The shaded area is between x=0 and x=3.\" width=\"250\" height=\"235\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 7.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q217550\">Hint<\/span><\/p>\n<div id=\"q217550\" class=\"hidden-answer\" style=\"display: none\">The two curves intersect at [latex]x=1.[\/latex]<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793262779\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793262779\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793262779\">[latex]\\frac{5}{3}[\/latex] units<sup>2<\/sup><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/cm88bTFvRU4?controls=0&amp;start=1269&amp;end=1353&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266835\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266835\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.1AreaBetweenCurves1269to1353_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.1 Area Between Curves&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1163\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 2. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":3,"template":"","meta":{"_candela_citation":"{\"2\":{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\"}}","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1163","chapter","type-chapter","status-publish","hentry"],"part":1160,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1163","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":2,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1163\/revisions"}],"predecessor-version":[{"id":1374,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1163\/revisions\/1374"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/1160"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1163\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1163"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1163"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1163"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1163"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}